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It is useful to point out that functions of the form f ( x ) = 1 / ( x a ) n , where n is a positive integer, have infinite limits as x approaches a from either the left or right ( [link] ). These limits are summarized in [link] .

Two graphs side by side of f(x) = 1 / (x-a)^n. The first graph shows the case where n is an odd positive integer, and the second shows the case where n is an even positive integer. In the first, the graph has two segments. Each curve asymptotically towards the x axis, also known as y=0, and x=a. The segment to the left of x=a is below the x axis, and the segment to the right of x=a is above the x axis. In the second graph, both segments are above the x axis.
The function f ( x ) = 1 / ( x a ) n has infinite limits at a .

Infinite limits from positive integers

If n is a positive even integer, then

lim x a 1 ( x a ) n = + .

If n is a positive odd integer, then

lim x a + 1 ( x a ) n = +

and

lim x a 1 ( x a ) n = .

We should also point out that in the graphs of f ( x ) = 1 / ( x a ) n , points on the graph having x -coordinates very near to a are very close to the vertical line x = a . That is, as x approaches a , the points on the graph of f ( x ) are closer to the line x = a . The line x = a is called a vertical asymptote    of the graph. We formally define a vertical asymptote as follows:

Definition

Let f ( x ) be a function. If any of the following conditions hold, then the line x = a is a vertical asymptote of f ( x ) .

lim x a f ( x ) = + or −∞ lim x a + f ( x ) = + or −∞ or lim x a f ( x ) = + or −∞

Finding a vertical asymptote

Evaluate each of the following limits using [link] . Identify any vertical asymptotes of the function f ( x ) = 1 / ( x + 3 ) 4 .

  1. lim x −3 1 ( x + 3 ) 4
  2. lim x −3 + 1 ( x + 3 ) 4
  3. lim x −3 1 ( x + 3 ) 4

We can use [link] directly.

  1. lim x −3 1 ( x + 3 ) 4 = +
  2. lim x −3 + 1 ( x + 3 ) 4 = +
  3. lim x −3 1 ( x + 3 ) 4 = +

The function f ( x ) = 1 / ( x + 3 ) 4 has a vertical asymptote of x = −3 .

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Evaluate each of the following limits. Identify any vertical asymptotes of the function f ( x ) = 1 ( x 2 ) 3 .

  1. lim x 2 1 ( x 2 ) 3
  2. lim x 2 + 1 ( x 2 ) 3
  3. lim x 2 1 ( x 2 ) 3

a. lim x 2 1 ( x 2 ) 3 = ; b. lim x 2 + 1 ( x 2 ) 3 = + ; c. lim x 2 1 ( x 2 ) 3 DNE. The line x = 2 is the vertical asymptote of f ( x ) = 1 / ( x 2 ) 3 .

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In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points.

Behavior of a function at different points

Use the graph of f ( x ) in [link] to determine each of the following values:

  1. lim x −4 f ( x ) ; lim x −4 + f ( x ) ; lim x −4 f ( x ) ; f ( −4 )
  2. lim x −2 f ( x ) ; lim x −2 + f ( x ) ; lim x −2 f ( x ) ; f ( −2 )
  3. lim x 1 f ( x ) ; lim x 1 + f ( x ) ; lim x 1 f ( x ) ; f ( 1 )
  4. lim x 3 f ( x ) ; lim x 3 + f ( x ) ; lim x 3 f ( x ) ; f ( 3 )
The graph of a function f(x) described by the above limits and values. There is a smooth curve for values below x=-2; at (-2, 3), there is an open circle. There is a smooth curve between (-2, 1] with a closed circle at (1,6). There is an open circle at (1,3), and a smooth curve stretching from there down asymptotically to negative infinity along x=3. The function also curves asymptotically along x=3 on the other side, also stretching to negative infinity. The function then changes concavity in the first quadrant around y=4.5 and continues up.
The graph shows f ( x ) .

Using [link] and the graph for reference, we arrive at the following values:

  1. lim x −4 f ( x ) = 0 ; lim x −4 + f ( x ) = 0 ; lim x −4 f ( x ) = 0 ; f ( −4 ) = 0
  2. lim x −2 f ( x ) = 3 . ; lim x −2 + f ( x ) = 3 ; lim x −2 f ( x ) = 3 ; f ( −2 ) is undefined
  3. lim x 1 f ( x ) = 6 ; lim x 1 + f ( x ) = 3 ; lim x 1 f ( x ) DNE; f ( 1 ) = 6
  4. lim x 3 f ( x ) = ; lim x 3 + f ( x ) = ; lim x 3 f ( x ) = ; f ( 3 ) is undefined
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Evaluate lim x 1 f ( x ) for f ( x ) shown here:

A graph of a piecewise function. The first segment curves from the third quadrant to the first, crossing through the second quadrant. Where the endpoint would be in the first quadrant is an open circle. The second segment starts at a closed circle a few units below the open circle. It curves down from quadrant one to quadrant four.

Does not exist.

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Chapter opener: einstein’s equation

A picture of a futuristic spaceship speeding through deep space.
(credit: NASA)

In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object can travel. Given Einstein’s equation for the mass of a moving object, what is the value of this bound?

Our starting point is Einstein’s equation for the mass of a moving object,

m = m 0 1 v 2 c 2 ,

where m 0 is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at high speeds, we can graph the ratio of masses m / m 0 as a function of the ratio of speeds, v / c ( [link] ).

A graph showing the ratio of masses as a function of the ratio of speed in Einstein’s equation for the mass of a moving object. The x axis is the ratio of the speeds, v/c. The y axis is the ratio of the masses, m/m0. The equation of the function is m = m0 / sqrt(1 –  v2 / c2 ). The graph is only in quadrant 1. It starts at (0,1) and curves up gently until about 0.8, where it increases seemingly exponentially; there is a vertical asymptote at v/c (or x) = 1.
This graph shows the ratio of masses as a function of the ratio of speeds in Einstein’s equation for the mass of a moving object.

We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speed of light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at v / c = 1 . We can try a few values of this ratio to test this idea.

Ratio of masses and speeds for a moving object
v c 1 v 2 c 2 m m 0
0.99 0.1411 7.089
0.999 0.0447 22.37
0.9999 0.0141 70.71

Thus, according to [link] , if an object with mass 100 kg is traveling at 0.9999 c , its mass becomes 7071 kg. Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light.

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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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